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Count Symmetric Integers

Difficulty: Easy

Problem Description

You are given two positive integers low and high. An integer x consisting of 2 * n digits is symmetric if the sum of the first n digits of x is equal to the sum of the last n digits of x. Numbers with an odd number of digits are never symmetric. Return the number of symmetric integers in the range [low, high].

Key Insights

  • Symmetric integers must have an even number of digits.
  • The range for checking symmetric integers can be limited to even-length numbers.
  • The sum of the first half of digits must equal the sum of the second half.
  • Efficiently iterate through the range and check each number.

Space and Time Complexity

Time Complexity: O(d), where d is the number of digits in the range. Space Complexity: O(1), as we are using a fixed amount of additional space.

Solution

The approach involves iterating through each integer from low to high and checking if it is symmetric. For each number, we will convert it to a string to split it into two halves, then calculate and compare the sums of the digits in each half. We will count how many integers meet the symmetry condition.

Code Solutions

def countSymmetricIntegers(low, high):
    def is_symmetric(num):
        s = str(num)
        n = len(s) // 2
        return sum(int(s[i]) for i in range(n)) == sum(int(s[i]) for i in range(n, len(s)))

    count = 0
    for num in range(low, high + 1):
        if len(str(num)) % 2 == 0 and is_symmetric(num):
            count += 1
    return count
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